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QUESTION

An integer $n\geq1$ is called a \textit{Carmichael number} if $n$
is not a prime and $a^{n-1}\equiv1$ (modulo $n$) for all integers
$a$ such that gcd$(a,n)=1$. Throughout this question let $n$
denote a Carmichael number..

\begin{description}

\item[(i)]
Show that $n$ cannot be a power of 2.

\item[(ii)]
Let $p$ be an odd prime such that $p^\alpha$ divides $n$ with
$\alpha\geq1$. Show that $\alpha=1$ and $p-1$ divides $n-1$.
(Hint: You may need to use the Chinese Remainder Theorem to make a
good choice of $\alpha$. Also you may assume that the group of
units modulo $p^\alpha$, usually denoted by $U_{p^\alpha}$, is
cyclic of order $\phi(p^\alpha)$.)

\item[(iii)]
Use part (ii) to show that $n$ is odd.

\item[(iv)]
Show that $n=p_1p_2\ldots p_r$ where $p_1,\ldots,p_r$ are distinct
odd primes and $r\geq3$.

\end{description}



ANSWER

\begin{description}

\item[(i)]
Suppose that $n=2^\alpha$, Then $\alpha=1$ corresponds to $n=2$,
which is prime, while for $\alpha\geq2$ we may take $a=-1$ to
obtain

$$a^{2^\alpha-1}=(-1)^{2^\alpha-1}=-1$$

which is not congruent to 1 modulo 4 much less modulo $a^\alpha$.

\item[(ii)]
Write $n=p_1^{\alpha_1}\ldots p_r^{\alpha_r}$ with each
$\alpha_i\geq1$ and $p_1,\ldots p_r$ distinct primes. Suppose that
the odd prime $p$ is $p_1$ so that $\alpha=\alpha_1$. By the
Chinese Remainder Theorem we may choose an integer, $a$, such that

$$a=\left\{\begin{array}{cl}x&\textrm{(modulo
$p_1^{\alpha_1})$},\\1&\textrm{(modulo $p_i$) for $i=2,\ldots
r$}\end{array}\right.$$

where the multiplicative order of $x$ modulo $p_1^{\alpha_1}$ is
$\phi(p_1^{\alpha_1})=p_1^{\alpha_1-1}(p_1-1)$. Such a choice of
$a$ satisfies gcd$(a,n)=1$. Then $a^{n-1}\equiv1$ (modulo $n$)
implies that $x^{n-1}\equiv a^{n-1}\equiv1$ (modulo
$p_1^{\alpha_1}$). Therefore $p_1^{\alpha_1-1}(p_1-1)$ divides
$n-1=p_1^{\alpha_1}\ldots p_r^{\alpha_r}-1$. If $\alpha_1\geq2$
then $p_1$ would divide 1 so we must have $\alpha_1=1.$ In this
case the condition that $\phi(p_1^{\alpha_1})$ divides $n-1$
becomes simply $p_1==1$ divides $n-1$, as required.

\item[(iii)]
By part (i), $r\geq2$ in part (ii) and so one of the primes
dividing $n$ is odd. By part (ii). $n-1$ is even because it is
divisible by $p-1$ for some odd prime. Hence $n$ is odd.

\item[(iv)]
By parts (ii) and (ii), $n$ is a product of distinct odd primes.
It remains to show that $n=p_1p_2$ is not a carmichael number when
$p_1$ and $p_2$ are distinct primes. However the equation

$$n-1=p_1p_2-1=(p_1-1)(p_2+(p_2-1)$$

together with part (ii) would imply that $(p_2-1)$ divides
$(p_1-1)$ and vice versa. This would imply the contradiction that
$p_1=p_2$.

\end{description}




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